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Prove sqrt(xy) is less than or equal to (x+y)/2 for all positive values of x and y Watch

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    Prove sqrt(xy) is less than or equal to (x+y)/2 for all positive values of x and y
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    (Original post by Kartheyan)
    Prove sqrt(xy) is less than or equal to (x+y)/2 for all positive values of x and y
    Tried anything?
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    (Original post by Kartheyan)
    Prove sqrt(xy) is less than or equal to (x+y)/2 for all positive values of x and y
    Well if x,y positive then wlog x=a^2, y=b^2 means you want to prove 2ab <= a^2 + b^2.
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    (Original post by dhxrt)
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    It's a correct answer but you're depriving OP of some important experience by just posting the answer rather than hinting at it like other posters have done.
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    (Original post by I hate maths)
    It's a correct answer but you're depriving OP of some important experience by just posting the answer rather than hinting at it like other posters have done.
    I can explain the steps if needs, it’s just it’s a very strange question for proof that has quite an obscure answer that doesn’t come easily
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    (Original post by dhxrt)

    Sorry the formatting went strange but this is the answer ^^
    Posting full solutions is against the rules of this forum.
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    (Original post by dhxrt)
    I can explain the steps if needs, it’s just it’s a very strange question for proof that has quite an obscure answer that doesn’t come easily
    Where did you get (x-y)^2 from?
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    (Original post by dhxrt)
    I can explain the steps if needs, it’s just it’s a very strange question for proof that has quite an obscure answer that doesn’t come easily
    Kind of strange to see the AM-GM inequality described as obscure!
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    (Original post by Kartheyan)
    Where did you get (x-y)^2 from?
    So it comes from the fact that any number squared will result in a positive value (or zero if it is zero^2). It is convenient in this proof to start here
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    (Original post by DFranklin)
    Kind of strange to see the AM-GM inequality described as obscure!
    It is obscure given the fact that this is question very early on in the AS Maths textbook. I understand that it’s not difficult for many people, but I’ve seen lots struggle on this question at the start of AS
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    (Original post by dhxrt)
    So it comes from the fact that any number squared will result in a positive value (or zero if it is zero^2). It is convenient in this proof to start here
    I'm still a bit confused, where did the 4xy come from?
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    (Original post by DFranklin)
    Posting full solutions is against the rules of this forum.
    Oh I wasn’t aware and what’s the reason for that?
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    (Original post by Kartheyan)
    I'm still a bit confused, where did the 4xy come from?
    4xy is added to both sides so that the proof can continue, it’s just the next step of how you prove it.
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    (Original post by dhxrt)
    Oh I wasn’t aware and what’s the reason for that?
    Moderator power
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    (Original post by dhxrt)
    Oh I wasn’t aware and what’s the reason for that?
    Read the posting guidelines sticky at the top of the forum.
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    (Original post by Kartheyan)
    Where did you get (x-y)^2 from?
    You could start by going the other way and show that \displaystyle \sqrt{xy} \leq \frac{x+y}{2} \Rightarrow 0 \leq (x-y)^2 which is more natural, but then you would need to check your steps are reversible and the reverse implication is also correct since your proof needs a true statement to imply your required statement, not the other way around, as false statements can imply true statements.
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    (Original post by dhxrt)
    4xy is added to both sides so that the proof can continue, it’s just the next step of how you prove it.
    You only added it to the RHS.
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    (Original post by DFranklin)
    Read the posting guidelines sticky at the top of the forum.
    I stg this site needs to chill 😂
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    (Original post by Kartheyan)
    You only added it to the RHS.
    I didnt, if you look carefully at the LHS, the sign has changed: -2xy + 4xy = 2xy
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    (Original post by dhxrt)
    I stg this site needs to chill 😂
    Now you've been warned, I'll report further full solutions for removal.
 
 
 
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